成功 url 到当前 url

Question

我正在使用基于 Django 类的视图和表单。我希望视图在提交新消息后保持在同一页面上。网址就像

localhost:8000/chat/messages/username/

其中 'username' 是接收消息的用户的用户名

class ThreadView(LoginRequiredMixin, FormMixin, DetailView):
    template_name = 'thread.html'
    form_class = ComposeForm

    def get_queryset(self):
        return Thread.objects.by_user(self.request.user)

    def get_object(self):
        other_username  = self.kwargs.get("username")
        obj, created    = Thread.objects.get_or_new(self.request.user, other_username)
        if obj == None:
            raise Http404
        return obj

    def get_context_data(self, **kwargs):
        context = super().get_context_data(**kwargs)
        context['form'] = self.get_form()
        return context

    def post(self, request, *args, **kwargs):
        if not request.user.is_authenticated:
            return HttpResponseForbidden()
        self.object = self.get_object()
        form = self.get_form()
        if form.is_valid():
            return self.form_valid(form)
        else:
            return self.form_invalid(form)

    def form_valid(self, form):
        thread = self.get_object()
        user = self.request.user
        message = form.cleaned_data.get("message")
        ChatMessage.objects.create(user=user, thread=thread, message=message)
        return super().form_valid(form)

网址.py：

app_name = 'chat'

urlpatterns = [
    path("", InboxView.as_view()),
    url(r'^messages/(?P<username>[\w.@+-]+)', ThreadView.as_view(), name="chaturl"),
]

如果我给success_url='./'它去locahost:8000/chat/messages/，这会导致 404 错误。

Answer 1

您可以覆盖该get_success_url方法并反转网址：

from django.urls import reverse

def get_success_url(self):
    return reverse('chat:chaturl', kwargs={'username': self.kwargs['username']})

或者，如果您想要提交请求的路径，您可以使用request.path.

from django.urls import reverse

def get_success_url(self):
    return self.request.path